KCET · Maths · Inverse Trigonometric Functions
If \(2 \sin ^{-1} x-3 \cos ^{-1} x=4, x \in[-1,1]\), then \(2 \sin ^{-1} x+3 \cos ^{-1} x\) is equal to
- A \(\frac{4-6 \pi}{5}\)
- B \(\frac{6 \pi-4}{5}\)
- C \(\frac{3 \pi}{2}\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(\frac{6 \pi-4}{5}\)
Step-by-step Solution
Detailed explanation
\(\because 2 \sin ^{-1} x-3 \cos ^{-1} x=4\)
\(\Rightarrow \quad 5 \sin ^{-1} x=4+\frac{3 \pi}{2}\)
\(\therefore \quad \sin ^{-1} x=\frac{4+\frac{3 \pi}{2}}{5}=\frac{8+3 \pi}{10}\) ....(I)
So, \(2 \sin ^{-1} x+3 \cos ^{-1} x\)
\(=2 \sin ^{-1} x+3\left(\frac{\pi}{2}-\sin ^{-1} x\right)\)
\(=-\sin ^{-1} x+\frac{3 \pi}{2}\)
\(=\frac{-8-3 \pi}{10}+\frac{3 \pi}{2}=\frac{-8-3 \pi+15 \pi}{10}\)
\(=\frac{12 \pi-8}{10}=\frac{6 \pi-4}{5}\)
\(\Rightarrow \quad 5 \sin ^{-1} x=4+\frac{3 \pi}{2}\)
\(\therefore \quad \sin ^{-1} x=\frac{4+\frac{3 \pi}{2}}{5}=\frac{8+3 \pi}{10}\) ....(I)
So, \(2 \sin ^{-1} x+3 \cos ^{-1} x\)
\(=2 \sin ^{-1} x+3\left(\frac{\pi}{2}-\sin ^{-1} x\right)\)
\(=-\sin ^{-1} x+\frac{3 \pi}{2}\)
\(=\frac{-8-3 \pi}{10}+\frac{3 \pi}{2}=\frac{-8-3 \pi+15 \pi}{10}\)
\(=\frac{12 \pi-8}{10}=\frac{6 \pi-4}{5}\)
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